## Zonohedra and the Sylvester-Gallai theorem

When I was in Melbourne recently, I spent some time browsing through a copy of “Twelve Geometric Essays” by Harold Coxeter in the (small) library at AMSI. One of these essays was entitled “The classification of zonohedra by means of projective diagrams”, and it contained a very cute proof of the Sylvester-Gallai theorem, which I thought would make a nice (short!) blog post.

The Sylvester-Gallai theorem says that a finite collection of points in a projective plane are either all on a line, or else there is some line that contains exactly two of the points. Coxeter’s proof of this theorem falls out incidentally from an apparently unrelated study of certain polyhedra known as zonohedra.

For subsets $P$ and $Q$ of a vector space $V$, the Minkowski sum $P+Q$ is the set of points of the form $p+q$ for $p\in P$ and $q \in Q$. If $P$ and $Q$ are polyhedra, so is $P + Q$, and the vertices of $P+Q$ are sums of vertices of $P$ and $Q$. One natural way to think of $P+Q$ is that it is the projection of the product $P\times Q$ under the affine map $+:V\times V \to V$.

The simplest definition of a zonohedron (in any dimension) is that it is the Minkowski sum of finitely many intervals. Thus the faces of a zonohedra are themselves zonohedra. In 2 dimensions a zonohedron is a centrally symmetric polygon, and therefore has an even number of edges which come in parallel pairs of the same length. A zonohedron is convex, being the Minkowski sum of convex sets. Thus it is topologically a ball, and its boundary is topologically a sphere. A parallelepiped is an example of a 3-dimensional zonohedron; so is the rhombic dodecahedron and the rhombic triacontahedron. One can think of a zonohedron as a projection to a low dimensional space of a high dimensional parallelepiped; one can use this observation to produce interesting aperiodic tilings from zonohedra.

Here is Coxeter’s proof of the Sylvester-Gallai theorem. Let $Z= +_i I_i$ be a 3-dimensional zonohedron, expressed as the Minkowski sum of some collection of intervals $I_i$. Each $I_i$ determines a point $p_i$ in the projective plane; conversely, a collection of points in the projective plane determines a family of zonohedra, where each element of the family is determined by the edge lengths of the $I_i$. The faces of the zonohedra correspond to the colinear collections of $p_i$. A decomposition of the sphere into polygons meeting at least 3 to a vertex must contain at least one polygon with $<6$ sides, by Euler’s formula; hence every 3 dimensional zonohedron has at least one face with exactly $4$ sides. This corresponds to a line containing exactly 2 of the $p_i$; qed.

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### 5 Responses to Zonohedra and the Sylvester-Gallai theorem

1. Dylan Thurston says:

Wow, that’s really cute and beautiful.

2. Terence Tao says:

This sounds closely related to Melchior’s proof of Sylvester-Gallai:

http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem#Melchior.27s_proof

Here, one uses projective duality, to equate the problem to showing that given a bunch of projective lines, not all concurrent, there is a point incident to exactly two of these lines. The lines divide the projective plane into polygons, and one can then apply Euler’s formula, eventually concluding that the number of edges is less than three times the number of vertices, which implies that there is a vertex incident to just two lines.

• Danny Calegari says:

Hi Terry – that’s very nice – I had never heard of Melchior’s proof (isn’t Melchior one of the three wise men?). I think the relation is as follows: if $p_i$ is a collection of points, and $l_i$ is the dual collection of lines, the cellulation of the projective plane dual to the cellulation by $\cup_i l_i$ is combinatorially equivalent to the cellulation of the boundary of any associated zonohedron (here one must first take the double cover to get a cellulation of a sphere instead of a projective plane).

3. Kanal Temizleme says:

thanks